a pion at rest can decay into a muon and a neutrino. Conservation of energy and 3-momentum require
Conservation of energies $$E_\pi = E_\mu + E_\nu$$
Conservation of momentum $$P_\pi = P_\mu + P_\nu$$
but $$P_\pi = 0 $$ so $$P_\mu = – P_\nu$$
in this case we get
$$E_\pi = m_\pi c^2 $$
$$E_\mu = c \sqrt{m^2_\mu c^2+P^2_\mu}$$
and
$$ E_\nu = P_\nu c = P_\mu c $$
putting these in conservation of energy we get
$$ m_\pi c^2 = c \sqrt{m^2_\mu c^2+P^2_\mu } + P_\mu c $$
and solving for $P_\mu$ gives $$P_\mu = \frac{(m_\pi^2 -m^2_\mu)c}{(2m_\pi)}$$
we are given that $$E_\mu = \frac{(m_\pi^2+m^2_\mu)c^2}{(2m_\pi)}$$
Question. I just can't work out how the value of $E_\mu$ is obtained. If someone can explain that it would be wonderful.
Best Answer
It's just algebra.
You can start with $$p_\mu = \sqrt{E_\mu^2 - m_\mu^2} = \frac{m_\pi^2-m_\mu^2}{2m_\pi},$$ square both sides: $$E_\mu^2 - m_\mu^2 = \frac{m_\pi^4 + m_\mu^4 - 2m_\pi^2 m_\mu^2}{4m_\pi^2}, $$ rearrange: $$E_\mu^2 = \frac{m_\pi^4 + m_\mu^4 - 2m_\pi^2 m_\mu^2}{4m_\pi^2} + m_\mu^2 = \frac{m_\pi^4 + m_\mu^4 - 2m_\pi^2 m_\mu^2}{4m_\pi^2} + \frac{4m_\pi^2 m_\mu^2}{4m_\pi^2} = \frac{m_\pi^4 + m_\mu^4 + 2m_\pi^2 m_\mu^2}{4m_\pi^2}$$
then take the square root:
$$E_\mu=\frac{m_\pi^2+m_\mu^2}{2m_\pi}.$$